a, Ta có: \(n_{MgCO_3}=\dfrac{12,6}{84}=0,15\left(mol\right)\)
PT: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Theo PT: \(n_{CO_2}=n_{MgCO_3}=0,15\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,15.22,4=6,72\left(l\right)\)
b, Theo PT: \(n_{MgCl_2}=n_{MgCO_3}=0,15\left(mol\right)\)
Ta có: m dd sau pư = 12,6 + 50 - 0,15.44 = 56 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,15.95}{56}.100\%\approx25,45\%\)
c, Ta có: \(n_{KOH}=0,1\left(mol\right)\)
\(\Rightarrow\dfrac{n_{KOH}}{n_{CO_2}}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\)
→ Pư tạo muối KHCO3.
PT: \(KOH+CO_2\rightarrow KHCO_3\)
Theo PT: \(n_{KHCO_3}=n_{KOH}=0,1\left(mol\right)\)
⇒ mKHCO3 = 0,1.100 = 10 (g)
